Let -x- denote greatest integer less than or equal to x- If for n - 1-x - x3n - -j-03n ajxj- then -j-0-3n-2 -a2j - 4 -j-0-3n-1-2 - a2j - 1 is equal to-

(1 x + x^3)^n = ∑_j=0^3n a_jx^j (1 x + x^3)^n = a_0 + a_1 x + a_2 x^2 + .... + a_3n x^3n Put x = 1 1 = a_0 + a_1 + a_2 + a_3 nbsp;+ a_4 + ...+a_3n⋯(1) Put ...

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Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

Let [x] denot...

Question

Let $[x]$ denote greatest integer less than or equal to $x$ . If for $n \in N,$ $(1 - x + x^{3})^{n} = 3 n \sum j = 0 a_{j} x^{j}$ , then $[\frac{3 n}{2}] \sum j = 0 a_{2 j} + 4 [\frac{3 n - 1}{2}] \sum j = 0 a_{2 j + 1}$ is equal to:

1

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n

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2^{n - 1}

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Solution

The correct option is A $1$
$(1 - x + x^{3})^{n} = 3 n \sum j = 0 a_{j} x^{j} (1 - x + x^{3})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{3 n} x^{3 n}$
Put $x = 1$
$1 = a_{0} + a_{1} + a_{2} + a_{3} + a_{4} + . . . + a_{3 n} \dots (1)$
Put $x = - 1$
$1 = a_{0} - a_{1} + a_{2} - a_{3} + a_{4} . . . . (- 1)^{3 n} a_{3 n} \dots (2)$
From $(1) + (2)$
$\Rightarrow a_{0} + a_{2} + a_{4} + a_{6} + . . . = 1$
From $(1) - (2)$
$\Rightarrow a_{1} + a_{3} + a_{5} + a_{7} + \dots = 0$
Now
$[\frac{3 n}{2}] \sum j = 0 a_{2 j} + 4 [\frac{3 n - 1}{2}] \sum j = 0 a_{2 j + 1} = (a_{0} + a_{2} + a_{4} + \dots) + 4 (a_{1} + a_{3} + \dots) = 1 + 4 \times 0 = 1$

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Let [x] denote greatest integer less than or equal to x. If for n∈N, (1−x+x3)n=3n∑j=0ajxj, then [3n2]∑j=0a2j+4[3n−12]∑j=0a2j+1 is equal to:

Let $[x]$ denote greatest integer less than or equal to $x$ . If for $n \in N,$ $(1 - x + x^{3})^{n} = 3 n \sum j = 0 a_{j} x^{j}$ , then $[\frac{3 n}{2}] \sum j = 0 a_{2 j} + 4 [\frac{3 n - 1}{2}] \sum j = 0 a_{2 j + 1}$ is equal to: